PHOTOELECTRIC EFFECTS

QUANTUM PHYSICS

7.2 PHOTOELECTRIC EFFECT Part 1
CG-ANA-AF
SMK ABDUL RAHMAN TALIB, KUANTAN, PAHANG

IAUA session .

Electromagnetic radiations

Visible light

Range of λ. Longest? Shortest?

Range of f. Highest? Lowest?

Light as photons

Each photon carry energy, E

Which color has the highest energy?

What is power of photons?

What is intensity? Brightness?

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Gigi

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Michael - R
Jackson - O
Kena - Y
Halau - G
Balik - B

India - I
Utara - V

EM spectrum – 7 waves – highest λ/f – shortest λ/f – speed - infra – ultra – visible to eye – 7 colors combine

▪ Part 1 – What is PE ▪ Idea
▪ Part 2 – Examples of PE ▪ Activity 1 – Study PE

▪ Part 3 – Eintein’s PE equation ▪ Red photons VS Blue photons
▪ Part 4 – Photocells and Application ▪ Green photons VS Violet photons
▪ 4 Characteristics of PE effect
of PE ▪ Simulations
▪ Part 5 – Practice Makes Perfect
▪ Derivation
▪ Graphs

▪ Applications – Solar cells, Solar power plant, Light detectors,
Image sensor, Solar panels

▪ Objective & Structured Qs

▪ Photo ▪ Light = Photons ▪ Why is photoelectric
▪ Electric ▪ Metal effect important?
▪ Electrons
▪ How does it relate with
=Photoelectrons quantum physics?

Photoelectric effect is the……e…m…i…ss…io…n……….. of free
……el…ec…t…ro…n…s……….. from a metal surface when …l…ig…h…t………………..
strikes it.

It happens when a metal surface is …i…llu…m……in…a…te…d………….. by a
beam of light/electromagnetic radiation at a certain ……fr…e…qu…e…n…c…y ..,
electrons can be emitted from the ……m……et…a…l …………...

ill…um…in…ate…d
e…le…ctr…on…s

p…ho…to.e.l…ec…tro…ns

an.…ode p.…os…itiv.e ▪ Photons
▪ E = hf
cat.…ho…de .. an.…ode cu.…rre…nt
va.…lue ▪ (+)ve
▪ Accelerate PE

▪ Metal - e
▪ Work

function, W
▪ E → W+KE

COMPARE
▪ Metal
▪ Light Source
▪ Frequency of photons
▪ Energy of photons
▪ Current

▪ Cathode = Metal
▪ Illuminated with RED light

= photon
▪ No deflection = No current = No Pe-

▪ Cathode = Metal
▪ Illuminated with BLUE light

= photon
▪ Deflection = Current = Pe-





(1) Photoelectric effect occur
at certain frequency

- Emission of photoelectrons occurs at
certain frequency of radiation.

- Eviolet > W lithium = PE emitted
- E green < W lithium = NO PE emitted

Classical Theory
- Continuous energy
- Long exposure to light –PE should be

emitted at any light frequency

Low frequency but higher
intensity

- PE occurs at certain frequency
- Even if the INTENSITY increases

- NO PE emitted

Classical Theory
- Energy depends on intensity
- Bright light – High energy – Emit PE quickly
- Dim light – Low energy – Longer time to

absorb enough energy – Emit

(2) Threshold frequency
(3) Instantaneous effect

- If Eviolet > W lithium = PE emitted
- If f > f0 = PE emitted
- The emission of photoelectrons is

instantaneous even if in low intensity
photons. No time delay

Classical Theory
- Radiation energy is spread over the

wavefront
- Electron must gather sufficient energy
- Time lapse b/w irradiation and emission

(4) Intensity and Max Kinetic
Energy of Photoelectrons

- Intensity ↑ = photons ↑ illuminated on the
surface

- PE ↑ are ejected from the lithium surface
- Intensity doesn’t affect the Kinetic energy of

PE
- KE ∝ f , the higher the f, the higher the KE

Classical Theory

https://phet.colorado.edu/sims/cheerpj/photoelectric/latest/photoelectric.html?simulation=photoelectric

1. Photoelectric effect occurs when light strikes the surface of a metal.

2. The electrons in the metal absorb energy from the light and escape from the metal surface.

3. According to classical theory, light in wave form is a spectrum with continuous energy and photoelectric effect
should be able to occur at any light wave frequency.

4. Bright light which has high energy should be able to emit electrons quickly.

5. Dim light has low energy, so the electrons need a longer time to absorb enough energy to escape from the metal
surface.

6. However, the results of the photoelectric effect experiments show that the emission of photoelectrons only
apply to light waves with frequencies that exceed a certain value without being affected by the intensity of the
light.

7. Photoelectrons are also emitted instantaneously at those light frequencies even at low light intensities

(a) the effect of frequency on the photoelectric effect
Emission of photoelectrons occurs at certain frequency of radiation.

(b) the existence of a threshold frequency
If the frequency of the incident photon is greater than the threshold frequency of the
metal, electrons will emit from the surface.

The electrons that are emitted from the metal surface is called photoelectrons.
(c) the instantaneous emission of photoelectrons when light shines on it

If frequency of photon greater than threshold frequency, the emission of photoelectrons
is instantaneous – with no time delay even if the intensity is low. 1TO1.

(d) the effect of the intensity of light on the kinetic energy of photoelectron
▪ Increasing the intensity of light, increase the number of photons in one second
▪ This will increase number of photoelectrons not the energy to each PE
▪ The Kmax is not dependent on the intensity of light on the metal

QUANTUM PHYSICS

7.2 PHOTOELECTRIC EFFECT Part 2 – Discuss in Questions
- Determine Planck’s constant

THE PLANCK’S CONSTANT

DSKP – Carry out activity to determine the value of Planck’s constant using the Planck’s constant kit.
We can determine Planck's constant h by exposing a photocell to monochromatic light, i.e., light of a
specific wavelength, and measuring the kinetic energy EKE of the ejected electrons.

LED = Light Emitting Diode
▪ Photocell VS LED
▪ Photocell – light produces current
▪ LED – reverse photoelectric effect device
▪ LED – current produces light
▪ Both can determine Planck’s constant

ENERGY CONVERSION
▪ Electric potential energy »» Light Energy
▪ EPE = eVa
▪ Light = Photon = hf = hc/λ

Simulation source:
http://mpv-au.vlabs.ac.in/modern-
physics/Determination_of_Plancks
_Constant/experiment.html

Arrangement of Apparatus

Tabulation of Data

LED COLOR λ (nm) λ (m) Activation Voltage, 1/λ (m⁻¹) Planck's constant, h =
RED 650 Va (V) #DIV/0! eλVa/c (Js)
570
YELLOW 510 1.908
GREEN 467
BLUE 2.178

2.434 #DIV/0!
2.662

2.7 (1.538 x 106, 1.908)
(2.14 x 106, 2.66) (1.754 x 106, 2.178)
(1.961 x 106, 2.434)
2.6 (2.141 x 106, 2.662)

2.5 Find Gradient, m
2.4 Gradient, m = . (2.66 – 1.91) .

2.3 (2.14 – 1.54) x 106
2.2 Gradient, m = .1.25 x 10-6 Vm

2.1
2.0
1.9

(1.54 x 106, 1.91)
1.8

1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2

2.7 ENERGY CONVERSION
(2.14 x 106, 2.66)
▪ eVa = h c
2.6 λ

2.5 ▪ Rearrange the formula : Va = h c . 1
2.4 eλ

2.3 ▪ From Va – 1 ; gradient, m = h c
2.2 λe

2.1 ▪ Hence, h = me
2.0 c
1.9
▪ Hence, h = (1.25 x 10-6)(1.6 x 10-19)
(1.54 x 106, 1.91) (3.0 x 108)
1.8
▪ Hence, h = 6.67 x 10-34 Js
1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2

QUANTUM PHYSICS

7.3 EINSTEIN’S PHOTOELECTRIC EQUATION Part 1
CG-ANA-AF
SMK ABDUL RAHMAN TALIB, KUANTAN, PAHANG

7.3 EINSTEIN’S PHOTOELECTRIC THEORY

Learning Standards:
(1)State minimum work function needed by a metal to emit an electron using

Einstein’s equation
(2)Explain threshold frequency, fo and work function, W
(3)Determine work function of metal, W=hfo
(4)Solve problems involving Einstein’s equation for photoelectric effect,

hf = W + ½ mv2

Photon, E 1TO1 Photon energy, E = Minimum energy required + Maximum kinetic
to release a photoelectron energy of a
photoelectron

Photoelectron, E = W + Kmax
Kmax hf = W + ½ mvmax2

Metal, W = hf0

Minimum energy Maximum kinetic
Photon energy, E = required to release + energy of a
photoelectron
a photoelectron

E = W + Kmax
hf = W + ½ mvmax2

If the energy of photon = W; hence Kmax = 0

hf = W + 0
hf0 = W ; f0 = threshold frequency

E = hf0 + ½ mvmax2

Minimum energy Maximum kinetic
Photon energy, E = required to release + energy of a
photoelectron
a photoelectron

E = W + Kmax

Work Function, W = Minimum ENERGY required to release a photoelectron
Threshold frequency, f0 = Minimum FREQUENCY required to release a photoelectron

Photoelectric effect exists Derive Formula:

f0

Graph of Kmax against f

W

Work Function, W = Minimum ENERGY required to release a photoelectron
Threshold frequency, f0 = Minimum FREQUENCY required to release a photoelectron

Photoelectric effect exists

f0 f0 B f0 A f0 C

Graph of Kmax against f Graph of Kmax against f for different types of metals

Threshold wavelength, λ0 = Maximum WAVELENGTH OF PHOTON required to
release a photoelectron

Photoelectric effect exists Derive Formula:

1/λ0 1/λ

Graph of Kmax against 1/λ

Photon, 1TO1 E = W +Kmax
E hf = W +Kmax
Photoelectron,
Kmax = hf0 + ½ mvmax2
Work function ➔ Metal
W

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Emission of electrons from a metal surface when shone on by light of a certain
frequency

• Yes.
• Bright light means more photons in a second
• The number of photoelectrons emitted is higher

(a) the effect of frequency on the photoelectric effect
Emission of photoelectrons occurs at certain frequency of radiation.

(b) the existence of a threshold frequency
If the frequency of the incident photon is greater than the threshold frequency of the
metal, electrons will emit from the surface.

The electrons that are emitted from the metal surface is called photoelectrons.
(c) the instantaneous emission of photoelectrons when light shines on it

If frequency of photon greater than threshold frequency, the emission of photoelectrons
is instantaneous – with no time delay even if the intensity is low. 1TO1.

(d) the effect of the intensity of light on the kinetic energy of photoelectron
▪ Increasing the intensity of light, increase the number of photons in one second
▪ This will increase number of photoelectrons not the energy to each PE
▪ The Kmax is not dependent on the intensity of light on the metal

• Light consists of discrete energy packets
• When a photon hits a metal surface, all its energy will be transferred to an

electron in the metal.
• If the frequency of light is higher than the threshold frequency of the metal the

photoelectron will be emitted instantaneously from the metal surface

• No.
• The intensity of light only affects the number of photons arriving on the metal

per second (photon rate).
• The maximum kinetic energy of a photoelectron is influenced by the photon

energy.
• Increasing the light intensity will not increase the kinetic energy of the

photoelectrons

h = 6.63 x 10-34 Js hf = W + Kmax
f = 6.67 x 1014 Hz (6.63 x 10-34) (6.67 x 1014) = (3.43 x 10-19) + Kmax
W = 4.42 x 10-19 J
Kmax = 4.42 x 10-19 - 3.43 x 10-19
f0 = - = 9.92 x 10-20 J
Kmax = ?

h = 6.63 x 10-34 Js W = hf0
= (6.63 x 10-34) (5.6 x 1014)
f =-
= 3.71 x 10-19 J
W =?
f0 = ? = 5.6 x 1014 Hz
Kmax = ?

h = 6.63 x 10-34 Js W = hf0
= (6.63 x 10-34) (9.85 x 1014)
f0 = 9.85 x 1014 Hz
W =? = 6.53 x 10-19 J

h = 6.63 x 10-34 Js hf = W + Kmax
f = 5.34 x 1014 Hz (6.63 x 10-34) (5.34 x 1014) = (3.43 x 10-19) + Kmax
W = 3.34 x 10-19 J
Kmax = ? Kmax = (6.63 x 10-34) (5.34 x 1014) - 3.43 x 10-19
= 1.1042 x 10-20 J

hc = 1.243 x 103 eV nm hf = W + Kmax
λ = 550 nm hc/ λ = W + Kmax
W = 2.00 eV 1.243 x 103/550 = 2.00 + Kmax

Kmax = ? = ½ mv2 Kmax = 0.26 eV
m = 9.11 x 10-31 kg = (0.26 x 1.6 x 10-19)
= 4.16 x 10-20 J

Kmax = ½ mv2
4.16 x 10-20 = ½ (9.11 x 10-31)v2

v = 3.02205 x 105 ms-1

hc = 1.243 x 103eV nm hf = W + Kmax
λ = 600 nm hc/ λ = W + Kmax
1.243 x 103/600 = 1.00 + Kmax
W = 1.00 eV
2.07 = 1.00 + Kmax
Kmax = ?
Kmax = 1.07 eV

To stop most energetic photoelectron is to supplied electric
potential energy with the same magnitude of Kmax

eV = Kmax
V = Kmax / e

= 1.07eV/e = 1.07 V

DO THIS FIRST = ANALYSE THE QUESTION!!!!

h = 6.63 x 10-34 Js hf = W + Kmax ……(1)
λr = 750 nm hc/ λ = W + Kmax ……(2)
λb = 460 nm hc/460nm = W + 2K ……(3)
hc/750nm = W + K
KmaxR = K 2hc/750nm = 2W + 2K

KmaxB = 2 KmaxR = 2K (3) – (2) 2hc/750nm = 2W + 2K ……(3)
hc/460nm = W + 2K ……(2)
W =?
2hc/750nm - hc/460nm = W
f0 = ? W = 9.8 x 10-20 J

(b) What is the threshold wavelength W = 9.8 x 10-20
of the photoelectric material? hc/λ = 9.8 x 10-20

λ = 0.000002029 = 2.029 x 10-6 m = 2029 nm

QUANTUM PHYSICS

7.3 EINSTEIN’S PHOTOELECTRIC EQUATION Part 2
CG-ANA-AF
SMK ABDUL RAHMAN TALIB, KUANTAN, PAHANG

7.3 EINSTEIN’S PHOTOELECTRIC

THEORY

DSKP:
(1)Explain production of photoelectric current in a photocell circuit
(2)Describe applications of photoelectric effect

PHOTOCELL ▪ Cathode = Lithium

▪ Diagram shows a photocell connected in ▪ Work function, W = 2.95 eV
series to a battery and microammeter.
▪ RED photon = Deflection of μA =
▪ The cathode is made of a material called Current = No PE
Lithium
▪ E < W; f < fo
▪ Work function of lithium is 2.95 eV.
▪ BLUE photon = Deflection of μA =
▪ When blue light is shined on the cathode, the Current = PE
meter shows a reading indicating there is a
current flowing through the photocell. ▪ E > W; f > fo

▪ When red light is shined onto the metal
cathode, the ammeter reading is zero. This
indicates that there is no current flowing
through the photocell.

▪ A photocell circuit consisting of (i) a glass vacuum tube; (ii) a semi-cylindrical (cathode) and (iii) metal rod

(anode)
▪ The semi-cylindrical cathode is (i) coated with a light-sensitive metal and (ii) connected to the negative potential.
▪ The anode is (i) a metal rod fixed at the axis of the semi cylindrical cathode and (ii) connected to the positive

potential.
▪ When the photocell is illuminated by light, the production of photoelectric current is produced in the circuit.

▪ Cathode = Lithium ▪ Cathode = Lithium

▪ Work function, W = 2.95 eV ▪ Work function, W = 2.95 eV

▪ RED photon = Deflection of μA = ▪ BLUE photon = Deflection of μA =
Current = No PE Current = PE

▪ E < W; f < fo ▪ E > W; f > fo